# WORD PROBLEMS TO FIND AREA OF QUADRILATERALS WITH VERTICES

Word Problems to Find Area of Quadrilaterals with Vertices :

Here we are going to see, some practice word problems to find the area of quadrilateral using given vertices.

## How to Find Area of Quadrilateral with Coordinates

To find the area of the quadrilateral with the given four vertices, we may use the formula given below.  ## Word Problems to Find Area of Quadrilaterals with Vertices - Questions

Question 1 :

Let P(11,7) , Q(13.5, 4) and R(9.5, 4) be the mid- points of the sides AB, BC and AC respectively of triangle ABC . Find the coordinates of the vertices A, B and C. Hence find the area of triangle ABC and compare this with area of triangle PQR

Solution :

To find the vertices of the triangle from the midpoint of the sides, please visit the page "https://www.onlinemath4all.com/how-to-find-the-vertices-of-a-triangle-if-the-midpoints-are-given.html" Vertex A :

=  (11 + 9.5 - 13.5, 7 + 4 - 4)

=  A (7, 7)

Vertex B :

=  (11 + 13.5 - 9.5, 7 + 4 - 4)

=  B (15, 7)

Vertex C :

=  (13.5 + 9.5 - 11, 4 + 4 - 7)

=  C (12, 1)

Area of triangle ABC : =  (1/2)[(49 + 15 + 84) - (105 + 84 + 7)]

=  (1/2)[148 - 196]

=  48/2

Area of triangle ABC  =  24 square units

Area of triangle PQR :

P(11,7) , Q(13.5, 4) and R(9.5, 4) =  (1/2)[(44 + 54 + 66.5) - (94.5 + 38 + 44)]

=  (1/2)[164.5 - 176.5]

=  (1/2) (12)

Area of triangle PQR  =  6

Area of triangle ABC  =  4 (Area of triangle PQR)

Let us look into the next example on "Word Problems to Find Area of Quadrilaterals with Vertices".

Question 2 :

In the figure, the quadrilateral swimming pool shown is surrounded by concrete patio. Find the area of the patio. Solution :

To find the area of patio, we have to subtract area of EFGH from area of ABCD.

Area of ABCD : =  (1/2)[(16 + 80 + 36 + 80) - (-64 - 24 - 100 - 24)]

=  (1/2)[(212)-(-212)]

=  (1/2)[212+212]

=  212 square units

Area of EFGH : =  (1/2)[(6 + 42 + 12 + 30) - (-30 - 6 - 42 - 12)]

=  (1/2)[(90)-(-90)]

=  (1/2)[90+90]

Area of EFGH  =  90 square units

Area of patio  =  212 - 90

=  122 square units.

Question 3 :

A triangular shaped glass with vertices at A(-5,-4) , B(1,6) and C(7,-4) has to be painted. If one bucket of paint covers 6 square feets, how many buckets of paint will be required to paint the whole glass, if only one coat of paint is applied.

Solution :

Area of triangle ABC  = =  (1/2)[(-30 - 4 - 28) - (-4 + 42 + 20)]

=  (1/2)[-62 - (58)]

=  (1/2)[-62 - 58]

=  (1/2)(120)

=  60 square feet

Area covered by one bucket of paint  =  6 square feets

Required number of bucket  =  60 / 6

=  10 buckets

Question 4 :

In the figure, find the area of (i) triangle AGF (ii) triangle FED (iii) quadrilateral BCEG Solution :

(i) triangle AGF =  (1/2) [(-2.5 - 13.5 - 6) - (-13.5 - 1 - 15)]

=  (1/2) [(-22) - (-29.5)]

=  (1/2) (-22+29.5)

=  (1/2)[7.5]

=  3.75 square units.

(ii) triangle FED

F (-2, 3) E (1.5, 1)  and D (1, 3) =  (1/2)[(-2 + 4.5 + 3) - (4.5 + 1 - 6)]

=  (1/2)[(5.5) - (-0.5)]

=  (1/2)6

Area of triangle FED  =  3 square units.

(iii) B (-4, -2) C (2, -1) E (1.5, 1) G (-4.5, 0.5) =  (1/2)[(4 + 2 + 0.75 + 9) - (-4 - 1.5 - 4.5 - 2)]

=  (1/2)[(15.75) - (-12)]

=  (1/2)(15.75 + 12)

=  (1/2)(27.75)

=  13.875 square units. After having gone through the stuff given above, we hope that the students would have understood, "Word Problems to Find Area of Quadrilaterals with Vertices".

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